2023 AMC 8 Problem 18
Below is the video solution and professionally curated solution for Problem 18 of the 2023 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 8 solutions, or check the answer key.
All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).
Difficulty rating: 1740
18.
Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump pads to the right or pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located pads to the right of her starting position?
Video solution:
Click to load, then click again to play
Written solution:
Let be the number of right jumps and be the number of left jumps. We need Modulo , this gives , so .
To minimize the total number of jumps, use the smallest possible , namely . Then , so .
The fewest number of jumps is .
Thus, D is the correct answer.
Problem 18 in Other Years
1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 AMC 8 · 1990 AMC 8 · 1991 AMC 8 · 1992 AMC 8 · 1993 AMC 8 · 1994 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1998 AMC 8 · 1999 AMC 8 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8